Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$. 
Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$

I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$
So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. But i have no idea how to prove this.
Something i’m not sure whether it’s true:
$$3(2^m)+4=n^2 \iff3(2^m+1)=n^2-1 \\ \iff 3(2^m+1)=(n+1)(n-1)$$
Thus $3\mid (n+1)(n-1) $, But since $3$ is a prime and $\gcd(n+1,n-1)=1$ or $2 \implies$
$3\mid n+1$ or $3\mid n-1$ What’s next?
 A: In respect of the stated comment, I add our second case (and so the whole solution) I left for OP to solve.
First, check $m=1$.
If $m≥2, m-2=k$ and $n=2n_1$, then
$$3×2^{k}+1=n_1^2, n_1≥2$$
Case $-1.$
$n_1=3a+1$, where $a≥1.$
$$3×2^k=9a^2+6a$$
$$2^k=3a^2+2a$$
$$a≥1 \Longrightarrow \begin{cases} a=2b \\ k≥3 \end{cases}$$
$$2^k=12b^2+4b$$
$$2^{k-2}=3b^2+b=b(3b+1)$$
$$b=2^x, 3b+1=2^y$$
If $x≥1$, then we get $3b+1≥7$ and $3b+1$ is an odd number, which gives a contradiction.
So, we deduce that $x=0.$
Then, we can check $n_1=3a+2$ by the same way.
Of course, $n_1=3a$ is impossible.

The method of looking at our second case is the same as the first one, as I mentioned earlier.
Case $-2.$
$n_1=3a+2$, where $a≥0.$
First, check $a=0$. If, $a≥1$ then
$$3×2^k+1=9a^2+12a+4$$
$$2^k=3a^2+4a+1, a≥1$$
$$a≥1 \Longrightarrow \begin{cases} a=2b-1 \\ k≥3 \end{cases}$$
We have,
$$2^k=3(2b-1)^2+4(2b-1)+1, b≥1$$
$$2^k=12b^2-4b$$
$$2^{k-2}=3b^2-b=b(3b-1)$$
$$b=2^x, 3b-1=2^y$$
If $x≥1$, then we get $3b-1≥5$ and $3b-1$ is an odd number, which gives a contradiction.
So, we deduce that $x=0.$
Finally, backward all the steps and you will find all required integer values ​​of $m$ and $n.$
A: (Here's a suggested approach. If you're stuck, explain where you're stuck and what you've tried.)

*

*Factor it as $ 3 \times 2^m = (n-2) ( n+2)$.

*Hence $ \{ n-2, n+2 \} = \{ 3 \times 2^a, 2^b \}$.

*Hence $  3 \times 2^a = 2^b \pm 4 $.

*If $ b = 0, 1, 2$, what can we say? What are the solutions with these small cases?


 $3 \times 2^a = 1+4, 2+4, 4+4$ (cannot take negative branch).
 So we get $ b=1, a = 1 \Rightarrow 2 \times 6 = 3\times 2^2  $.


*

*If $ b \geq 3$, what can we say?


 RHS is a multiple of 4 but not 8, so $ a = 2$.
$3 \times 4 = 2^b \pm 4$, which gives us $b=3, 4$.
$a = 2, b = 3 \Rightarrow 8\times 12 = 3\times 2^5 .$
$a = 2, b = 4 \Rightarrow 12 \times 16 = 3\times 2^6 . $

In conclusion, we have solutions only when $ m = 2, 5, 6$.
A: My english is not that good, I will try to explay the main steps, I hope you can understand.
$3(2^m)=(n^2-4)=(n-2)(n+2)$
so $3|n-2$ or $3|n+2$.
If $3|n-2$, then $n=3x+2$ for some $x\geq0$.
So, going back to the first equation with $n=3x+2$, we have $2^m=x(3x+4)$. So
$x=2^k$, for some $k$ satisfying $0\leq k \leq m$.
This means that. we have $2^{k-m}=3.2^k+4$, and so,
$3.2^k=4(2^{k-m-2}+1)$.
Finally $3.2^{k-2}=2^{k-m-2}+1$. And you analyse the conditions to both sides be odd or even.
The same idea, you can work with the case 3|n+2.
A: hint
It is easy to check that $ m $ satisfies
$$m\ge 2.$$
but
$$3.2^m=(n+2)(n-2)$$
with $ n $ even.
So
$$(n+2)|3.2^m $$
and by Gauss Theorem
$$(n+2)|2^m$$
